If any of you can do these problems, I would appreciate your help. I started finding curvature but i didn't know how to find the curvature with just the vector equation. I thought i needed y = f(x) equation to find curvature. Can anybody answer this by doing number 1 and so on?

Why was this in "Linear and Abstract Algebra" instead of "Calculus"? Actually, it would be better in the "homework help" section. Also we expect you to show what you have done on the problems yourself so that we can see what kind of help you need.

Q) A particle's position at time t is determined by the equation of the helix

r(t) = < cost, sint, t>

Let P be a point with a coordinates (0,1,pie/2).

1. Find the curvature k(t) at any time t.

Surely there are formulas in your text book for this. They involve both the first and second derivatives of the vector function. Look for them! By the way, you say "I thought i needed y = f(x) equation to find curvature." Well, you have r= f(t) which is what you need. This is a three dimensional problem and you cannot write "y= f(x)" for a three dimensional problem.

2. Find the center of the osculating circle at P.

The osculating circle is defined as the circle passing through P, in the plane containing both the tangent and normal vectors to the curve whose radius is 1 over the curvature you just found in 1.

3. Write down teh equation of teh osculating plane and the sphere whose intersection with that plane is the osculating circle.

The osculating plane is plane containing the point P and both the tangent and normal vectors at P.

4. Find a parametric equation of the osculating circle at P.

You found the center in 2 and you know the radius is 1 over the curvature you found in 1.

The acceleration vector itself is the second derivative of r(t). The "acceleration normal vector" is the component in the direction of the normal to the curve and the "tangential acceleration vector" is the component in the direction of the tangent to the curve.

Yes, i found the equation k(t) = (absolute value of r'(t) X r''(t)) / absolute value of r'(t) to the three. I found the k(t) to be 1/2.
I really can't solove number 2 using your explanation. You just defined what the osculating cirlce is. I realize that the raidus is reciprocal of teh curvature and the circle is in the palne containing both the tangent and normal vectors to the curve. But how do u find the center?.
And when you find the equation of the osculating plane and the sphere whose intersection with that plane is the osculating circle, which vector do u use for the slope of the equatin of the osculating plane.
Also, can you explane how to find a parametric equating of the osculating circle, using the center and the radius??????????

I haven't done any of this stuff in a while, so admittedly I would have to look up some things to do the problem. My advice to you is... look up some things. Spend some time thinking about what you already know (or can figure out) about equations of spheres and such. Really, you can probably do this without help, it doesn't sound that hard - trust your training, use the force, whatever. Good luck. Maybe someone else will be able to help you more, but nothing really takes the place of sweating over the problem awhile. In the end, your understanding will be better than it would be if we just told you how to do it. After you try a few more things, write again and I'll take a look at it. Good luck!
Aaron

First, please do not post the same question in several different places. The best place for this would have been in the "homework" section.

I'm glad you found the curvature (although it would be better to say "length of the vector" rather than "absolute value". Yes, the symbols are the same but at this point you are supposed to know the difference.). I presume you have also found the normal vector to the curve (it should be <0, -1, 0>). Since that vector already has length 1, just multiply by 1/curvature= 2 to get <0, -2, 0>. Start at the point, P= (0,1,&pi;/2), and "measure off" that distance in the direction of the normal vector: that is, add the vectors <0, 1, &pi/2> and <0,-2,0> to get <0, -1, &pi/2>, the "position vector" of the center. The center of the osculating circle is (0, -1, &pi/2).
Of course to write the parametric equations of the circle, you will have to know what plane it is in.

"And when you find the equation of the osculating plane and the sphere whose intersection with that plane is the osculating circle, which vector do u use for the slope of the equatin of the osculating plane."

I'm not sure what you mean by "slope of the equation".
Any plane, throught the point (x0,y0,z0) can be written in the form
A(x-x0)+ B(y-y0)+ C(z-z0)= 0.
Are the A, B, C, what you mean by "slope of the equation"? They are the x,y,z components of any vector normal to the plane. Since the plane contains both the tangent vector and normal vector to the curve, one way to find such a vector is to take the cross product of those two. I find that vector to be <1, 0, 1> and so the equation of the osculating plane is 1(x-0)+ 0(y-1)+ 1(z- &pi;/2)= 0 or
x+ z= &pi/2.

To find parametric equations of the osculating circle, here's how I would do it: You don't really need the equation of the osculating plane: the plane contains the tangent and normal vectors to the curve at P which are, by my calculations, <-1,0,1> and <0,-1,0>. Since those are themselves perpendicular, let's set up a coordinate system with those vectors giving the u, v directions. That is, u= 1, v=0, w= 0 is the same as x= -1/&radic;(2), y=0, z= 1/&radic;(2) (I divided by the length of the vector to get length 1) and when u= 0, v= 1, w= 0, x= 0, y= -1, v= 0. the vector perpendicular to both of those (we saw above) is <1, 0, 1> so we when u= 0, v=0, w= 1, we want x= 1/&sqrt;(2), y= 0, z= 1/&radic;(2). By making unit vectors correspond to unit vectors, the relationship must be linear:
u= ax+ by+ cz, etc. The conditions above give 9 equations for the 9 unknowns. Once you've found those, you can calculate that the coordinates of the center of the circle, in u, v, w, coordinates are (&pi;/(2&radic;(2)), -1, &pi;/(2&radic;(2)). Since the radius of the circle is still 2, we can write u= &pi;/(2&radic;(2))+ 2 cos(t),
v= -1+ 2 sin(t), w= &pi/(2&radic;(2)). Now convert back to x, y, z, coordinates: u= -x/&radic;(2)+ z/&radic;(2)= &pi;/(2&radic;(2))+ 2 cos(t), v= -y= -1+ 2 sin(t), w= x/&radic;(2)+ z/&radic;(2)= &pi/(2&radic;(2)). Solve those equations for x, y, z and you have your parametric equations.

Different ways of doing this: If you use "arclength" as parameter (so the derivative always has length 1: it IS the unit tangent vector), then the second derivative is alway perpendicular to the tangent vector- it IS the unit normal and its length is the curvature.

In this case, that's easy: you are given r(t)= (cos(t),sin(t),t) so
r'(t)= (-sin(t), cos(t),1). ||r'(t)||= &radic;(sin2(t)+ cos2(t)+ 1)= &radic;(2). Integrating, the arclength is given by s= &radic;(2)t so t= s/&radic;(2). That makes
r(s)= (cos(s/&radic;(2)),sin(s/&radic;(2),s/&radic;(2)),
r'(s)= (-1/&radic;(2)sin(s/&radic;(2)),1/&radic;(2)sin(s/&radic;(2)),1/&radic;(2)) which has length 1, and
r"= (-(1/2)cos(s/&radic;(2)), -(1/2)sin(s/&radic;(2)),0)=
(1/2)(-cos(s/&radic;(2)),- sin(s/&radic;(2)),0).
Yes, the curvature is 1/2.

In particular, when t= &pi;/2, s=&radic;(2)&pi;/2 so
r= (0, 1, &pi/2), r'= (-1, 0, 1), and r"= (0, -1/2, 0)
r' and r" are in the osculating plane so by finding a vector perpendicular to both, (1, 0, 1), we get that x+ z= &pi;/2 is the equation of the plane.

The sphere of radius 2 with center at (0, -1, &pi;/2) (the center we found for the osculating circle) would have parametric equations
x= 2cos(&theta;)sin(&phi;),
y= 2sin(&theta;)sin(&phi;)- 1, and
z= 2cos(&phi;)+ &pi;/2.

Since the osculating circle is in the plane x+ z= &pi/2;, we must have x+ z= 2cos(&theta;)sin(&phi;)+ 2cos(&phi;)+ &pi/2= &pi/2 or
cos(&theta)sin(&phi)= -cos(&phi;).

which means the parametric equations for the circle are
x= cos(&theta;)/ &radic(1+ cos2(&theta;)
In this case, that's easy: you are given r(t)= (cos(t),sin(t),t) so
r'(t)= (-sin(t), cos(t),1). ||r'(t)||= &radic;(sin2(t)+ cos2(t)+ 1)= &radic;(2). Integrating, the arclength is given by s= &radic;(2)t so t= s/&radic;(2). That makes
r(s)= (cos(s/&radic;(2)),sin(s/&radic;(2),s/&radic;(2)),
r'(s)= (-1/&radic;(2)sin(s/&radic;(2)),1/&radic;(2)sin(s/&radic;(2)),1/&radic;(2)) which has length 1, and
r"= (-(1/2)cos(s/&radic;(2)).
Yes, the curvature is 1/2.

In particular, when t= &pi;/2, s=&radic;(2)&pi;/2 so
r= (0, 1, &pi/2), r'= (-1, 0, 1), and r"= (0, -1/2, 0)
r' and r" are in the osculating plane so by finding a vector perpendicular to both, (1, 0, 1), we get that x+ z= &pi;/2 is the equation of the plane.

The sphere of radius 2 with center at (0, -1, &pi;/2) (the center we found for the osculating circle) would have parametric equations
x= 2cos(&theta;)sin(&phi;),
y= 2sin(&theta;)sin(&phi;)- 1, and
z= 2cos(&phi;)+ &pi;/2.

Since the osculating circle is in the plane x+ z= &pi/2;, we must have x+ z= 2cos(&theta;)sin(&phi;)+ 2cos(&phi;)+ &pi/2= &pi/2 or
cos(&theta)sin(&phi)= -cos(&phi;).